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Triginometry , algebra help

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data

    In △ABC, AB=(2-x) cm, BC=(x+1) cm and ∠ABC=120°:

    (a) Show that AC^2=x^2-x+7.

    (b) Find the value of x for which AC has a minimum value


    3. The attempt at a solution

    I am having trouble with part 'b'



    am suppose to complete the square , but I dont seem to get the right answer.

    [tex] x^2 -x +7 =0 [/tex]

    [tex] x^2 -x = -7 [/tex]

    [tex] (x- \frac{1}{2} )^2 = -7 + \frac{1}{4} [/tex]

    [tex](x- \frac{1}{2} )^2 = \frac{-27}{4} [/tex]

    [tex] x- \frac{1}{2} = -\frac{\sqrt27}{2}[/tex]


    and than am I suppose to solve for x ? what does it mean by minimum value?
     
  2. jcsd
  3. Mar 12, 2009 #2

    tiny-tim

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    uhh? :confused: where did "=0" come from??

    just minimise (x - 1/2)2 + 27/4 :wink:
     
  4. Mar 12, 2009 #3
    I dont understand what you mean by 'minimise' ? [tex] \frac{1}{4} + \frac{27}{4} = 7 [/tex]

    x=7?
     
  5. Mar 12, 2009 #4

    tiny-tim

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    Find the value of x for which (x - 1/2)2 + 27/4 is a minimum … what is that minimum? :smile:
     
  6. Mar 12, 2009 #5

    IS the minimum 27/4 , when x = 1/2?
     
  7. Mar 12, 2009 #6

    tiny-tim

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    Yup! :biggrin:

    That's what completing the square is all about! :wink:
     
  8. Mar 13, 2009 #7

    Mentallic

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    This step is not correct as taking the square root of a negative number (right hand side of equation) results in complex, imaginary numbers; not the negative square root as you've done. Since x is imaginary, it means that [tex]x^2-x+7 \neq 0[/tex]
    And of course, you've already seen that the minimum value is 27/4 which is more than 0.
     
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